Maths Questions

Find the minimum value of 𝑥^2 − 4𝑥 − 12.

-16

i) (53/4) -5(x-(3/2))²

ii) x=3/2

( 3/2 , 53/4 )

bruh do NOT let the 5 marks and big words scare u its so easy

ONLYYYY do iii (i only have the answer for that)

i didn’t really understand what bisect mean so it was confusing but now it makes sense:

When you bisect a line segment, you cut it into two equal lengths, so you want to find the midpoint

super easy just kinda forgot, it’s a disguised quadratic, use your own method not the one that is here, but just look at the answer

Remember, this isn’t a two‑number answer — it’s a range of values for k.

The discriminant becomes a quadratic expression in k, and a quadratic is positive only between its two roots.

So the original quadratic has real roots only when k lies between those two values.

but if you got it right.. well done! ;p

What does it mean if e.g. f(x) = -6 has no real solution

It doesnt intersect the curve

bruh easy question too how did i get it wrong, this why you dont rush!

What i did wrong was i made it so y=x/2 , but its actually y=(x/2) + 1 , remember 2/2 is 1 , it doesnt cancel out into 0

I was on this question for a while, I knewww that I knew how to do it but I just didn’t know where I was going wrong, turns out I factorised x²+k²x² like (2k²)x²not like (1 + k²)x² how I was meant to so yh

just do b

the centre is (4,-5) and radius is √40

How do we know which is correct?

We use the condition: “the gradient of AQ is positive.”

ez 6 marks

btw we letting u= -4-k

Calculate the value of each of the constants c and d (4 marks)

b) Complete the third row (2 marks)

c) State the limit suggested by Craig’s investigation for the gradient of these chords as h tends to 0 (1)

c) -5

f(x)=x−x^2

at x=3, because

f′(x)=1−2x

f′(3)=1−6=−5.

Using differentiation from first principles (4)

Hence, solve g(x) ≤ 0 (2 marks)